Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  long long t = 0, f = 1;
  char anss = getchar();
  while (!isdigit(anss)) {
    if (anss == '-') f = -1;
    anss = getchar();
  }
  while (isdigit(anss)) t = t * 10 + anss - '0', anss = getchar();
  return t * f;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    write(-x);
    return;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
inline void write_p(long long x) {
  write(x);
  putchar(' ');
}
const int N = 300005;
struct node {
  int l, len;
} p[N];
double sy, p1, p2;
int m, que, sum[N];
inline double Get_l(double x, double y) {
  if (x == p1) return p1;
  double k = (y - sy) / (x - p1);
  return x - y / k;
}
inline double Get_r(double x, double y) {
  if (x == p2) return p2;
  double k = (y - sy) / (x - p2);
  return x - y / k;
}
inline int Get(int l, int r) {
  if (l <= r)
    return sum[r] - sum[l - 1];
  else
    return 0;
}
inline int lower(int x) {
  int l = 1, r = m, ans = 0;
  while (l <= r) {
    int mid = l + r >> 1;
    if (p[mid].l <= x)
      ans = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  return ans;
}
inline void Solve(double x, double y) {
  double l = Get_l(x, y), r = Get_r(x, y);
  int tep = lower((int)l);
  int top = lower((int)r);
  if (tep == top) {
    if (p[tep].l <= l && p[top].l + p[top].len >= r) {
      double tans1 = r - l;
      printf("%.20lf\n", tans1 / y * (y - sy));
      return;
    }
  }
  double tans = Get(tep + 1, top - 1);
  if (1.0 * p[tep].l + 1.0 * p[tep].len > l)
    tans += (p[tep].l * 1.0 + p[tep].len * 1.0 - l);
  if (1.0 * p[top].l + 1.0 * p[top].len > r)
    tans += r - 1.0 * p[top].l;
  else if (p[top].l > l)
    tans += 1.0 * p[top].len;
  printf("%.20lf\n", tans / y * (y - sy));
}
int main() {
  sy = read();
  p1 = read();
  p2 = read();
  m = read();
  for (int i = 1; i <= m; ++i) p[i].l = read(), p[i].len = read() - p[i].l;
  for (int i = 1; i <= m; ++i) sum[i] = sum[i - 1] + p[i].len;
  que = read();
  for (int i = 1; i <= que; ++i) {
    int x = read(), y = read();
    Solve(x, y);
  }
}